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index of the original operator and the opposite sign. The same fact can be established directly by
the form of a regularizer (16) from the formula
˜
˜ a(t) – b(t) a(t)+ b(t) 1
˜
D(t)= = = .
˜
˜ a(t)+ b(t) a(t) – b(t) D(t)
Thus, for any singular operator with Cauchy kernel (15) of the normal type (a(t)±b(t) ≠ 0), there
exist infinitely many regularizers (16) whose characteristic part depends on an arbitrary function g(t)
˜
that contains an arbitrary regular kernel K(t, τ).
˜
Since the elements g(t) and K(t, τ) are arbitrary, we can choose them so that the regularizer
will satisfy some additional conditions. For instance, we can make the coefficient of ϕ(t)inthe
2
–1
2
regularized equation be normalized, i.e., equal to one. To this end we must set g(t)=[a (t)–b (t)] .
If no conditions are imposed, then it is natural to apply the simplest regularizers. These can be
˜
obtained by setting g(t) ≡ 1 and K(t, τ) ≡ 0 in formula (16), which gives the regularizer
b(t) ω(τ)
˜
∗◦
K[ω(t)] = K [ω(t)] ≡ a(t)ω(t) – dτ, (17)
πi L τ – t
1 b(τ) – b(t)
˜
or we can set g(t) ≡ 1 and K(t, τ)= – and obtain
πi τ – t
1 b(τ)ω(τ)
˜
K[ω(t)] = K [ω(t)] ≡ a(t)ω(t) – dτ. (18)
◦∗
πi L τ – t
The simplest operators K ∗◦ and K ◦∗ are most frequently used as regularizers.
Since the multiplication of operators is not commutative, we must distinguish two forms of
˜
regularization: left regularization, which gives the operator KK, and right regularization which
˜
leads to the operator KK. On the basis of the above remark we can claim that a right regular-
izer is simultaneously a left regularizer, and vice versa. Thus, the operation of regularization is
commutative.
˜
If an operator K is a regularizer for an operator K, then, in turn, the operator K is a regularizer
˜
for the operator K. The operators K 1 K 2 and K 2 K 1 can differ by a regular part only.
13.4-3. The Methods of Left and Right Regularization
Let a complete singular integral equation be given:
b(t) ϕ(τ)
K[ϕ(t)] ≡ a(t)ϕ(t)+ dτ + K(t, τ)ϕ(τ) dτ = f(t). (19)
πi L τ – t L
Three methods of regularization are used. The first two methods are based on the composition
of a given singular operator and its regularizer (left and right regularization). The third method
differs essentially from the first two, namely, the elimination of the singular integral is performed
by solving the corresponding characteristic equation.
1 . Left regularization. Let us take the regularizer (16):
◦
g(t)b(t) ω(τ)
˜
˜
K[ω(t)] ≡ g(t)a(t)ω(t) – dτ + K(t, τ)ω(τ) dτ. (20)
πi L τ – t L
˜
On replacing the function ω(t)in K[ω(t)] with the expression K[ϕ(t)] – f(t) we arrive at the
integral equation
˜
˜
KK[ϕ(t)] = K[f(t)]. (21)
˜
˜
By definition, KK is a Fredholm operator, because K is a regularizer. Hence, Eq. (21) is a Fredholm
equation. Thus, we have transformed the singular integral equation (19) into the Fredholm integral
equation (21) for the same unknown function ϕ(t).
This is the first regularization method, which is called left regularization.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

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